3.94.80 \(\int \frac {-15+30 x+45 \log (5)}{e^2} \, dx\)

Optimal. Leaf size=13 \[ \frac {15 x (-1+x+3 \log (5))}{e^2} \]

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Rubi [A]  time = 0.00, antiderivative size = 18, normalized size of antiderivative = 1.38, number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {9} \begin {gather*} \frac {15 (-2 x+1-3 \log (5))^2}{4 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-15 + 30*x + 45*Log[5])/E^2,x]

[Out]

(15*(1 - 2*x - 3*Log[5])^2)/(4*E^2)

Rule 9

Int[(a_)*((b_) + (c_.)*(x_)), x_Symbol] :> Simp[(a*(b + c*x)^2)/(2*c), x] /; FreeQ[{a, b, c}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {15 (1-2 x-3 \log (5))^2}{4 e^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 1.23 \begin {gather*} \frac {15 \left (-x+x^2+x \log (125)\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15 + 30*x + 45*Log[5])/E^2,x]

[Out]

(15*(-x + x^2 + x*Log[125]))/E^2

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fricas [A]  time = 1.01, size = 16, normalized size = 1.23 \begin {gather*} 15 \, {\left (x^{2} + 3 \, x \log \relax (5) - x\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((45*log(5)+30*x-15)/exp(2),x, algorithm="fricas")

[Out]

15*(x^2 + 3*x*log(5) - x)*e^(-2)

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giac [A]  time = 0.22, size = 16, normalized size = 1.23 \begin {gather*} 15 \, {\left (x^{2} + 3 \, x \log \relax (5) - x\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((45*log(5)+30*x-15)/exp(2),x, algorithm="giac")

[Out]

15*(x^2 + 3*x*log(5) - x)*e^(-2)

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maple [A]  time = 0.03, size = 15, normalized size = 1.15

method	result	size
gosper	\(15 x \left (x +3 \ln \relax (5)-1\right ) {\mathrm e}^{-2}\)	\(15\)
default	\({\mathrm e}^{-2} \left (45 x \ln \relax (5)+15 x^{2}-15 x \right )\)	\(20\)
risch	\(45 \ln \relax (5) {\mathrm e}^{-2} x +15 \,{\mathrm e}^{-2} x^{2}-15 x \,{\mathrm e}^{-2}\)	\(21\)
norman	\(15 \,{\mathrm e}^{-2} x^{2}+15 \left (3 \ln \relax (5)-1\right ) {\mathrm e}^{-2} x\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((45*ln(5)+30*x-15)/exp(2),x,method=_RETURNVERBOSE)

[Out]

15*x*(x+3*ln(5)-1)/exp(2)

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maxima [A]  time = 0.36, size = 16, normalized size = 1.23 \begin {gather*} 15 \, {\left (x^{2} + 3 \, x \log \relax (5) - x\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((45*log(5)+30*x-15)/exp(2),x, algorithm="maxima")

[Out]

15*(x^2 + 3*x*log(5) - x)*e^(-2)

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mupad [B]  time = 0.17, size = 15, normalized size = 1.15 \begin {gather*} \frac {{\mathrm {e}}^{-2}\,{\left (30\,x+45\,\ln \relax (5)-15\right )}^2}{60} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-2)*(30*x + 45*log(5) - 15),x)

[Out]

(exp(-2)*(30*x + 45*log(5) - 15)^2)/60

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sympy [A]  time = 0.05, size = 19, normalized size = 1.46 \begin {gather*} \frac {15 x^{2}}{e^{2}} + \frac {x \left (-15 + 45 \log {\relax (5 )}\right )}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((45*ln(5)+30*x-15)/exp(2),x)

[Out]

15*x**2*exp(-2) + x*(-15 + 45*log(5))*exp(-2)

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